18z2+54z+16=0

Solution for 18z2+54z+16=0 equation:

18z^2+54z+16=0

a = 18; b = 54; c = +16;
Δ = b2-4ac
Δ = 542-4·18·16
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-42}{2*18}=\frac{-96}{36}
=-2+2/3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+42}{2*18}=\frac{-12}{36}
=-1/3
$